trigonometry and bearings

[neelam]

A, B and C are points on a horizontal plane. A is 75m from C on a bearing of 210 and the bearing of B from C is 120. The bearing of B from A is 075. Find the distance BC. the question doesn't give a diagram and the answer is 75 m. but I don't have a clue how to work it out. So could someone please help, I've been staring at the question for the past half an hour and there's been no breakthrough!

 

[srmichael]

A, B and C are points on a horizontal plane. A is 75m from C on a bearing of 210 and the bearing of B from C is 120. The bearing of B from A is 075. Find the distance BC. the question doesn't give a diagram and the answer is 75 m. but I don't have a clue how to work it out. So could someone please help, I've been staring at the question for the past half an hour and there's been no breakthrough!

Start by placing C at the origin. From there, place A at a bearing of 210°. Do you know how to do that? Then place B at a bearing of 120°. Do this correctly and using the fact that the bearing of B from A is 75° will help you to determine BC.

Hint: You do not need to use any Law of Sines or Cosines for this one or any vector math. Just look at how the degrees fall into place and I think the answer will become apparent.

 

[neelam]

sorry - reply to trigonometry and bearing

sorry for disturbing you, your answer helped a lot but I'm slightly confused on what angle CAB is. I figured it would be 75degrees at first but that didn't work and then I tried 15degree (I did 360-210=135, then 135-75=60), but again that didn't work. sorry for asking again, If your not busy could you please reply. thankyou

 

[srmichael]

sorry for disturbing you, your answer helped a lot but I'm slightly confused on what angle CAB is. I figured it would be 75degrees at first but that didn't work and then I tried 15degree (I did 360-210=135, then 135-75=60), but again that didn't work. sorry for asking again, If your not busy could you please reply. thankyou

I wish I could draw a picture, but I am not as creative as others on this site so I will try and verbally explain.

Point A should be in quadrant III, 30° W of S and Point B should be in quadrant IV, 30° S of E. If you now picture Point A as a new origin, then Point B is at a bearing of 75° from Point A. If you draw this new y-axis through Point A perpendicular to the x-axis, call that point on the x-axis Point D. Therefore, angle DAB is now 75° with angle DAC being 30° (do you see why?) and thus angle CAB being 45°. What is the angle BCA??? Once you get that angle, the length of BC should be pretty straight-forward.

 

[DrPhil]

sorry for disturbing you, your answer helped a lot but I'm slightly confused on what angle CAB is. I figured it would be 75degrees at first but that didn't work and then I tried 15degree (I did 360-210=135, then 135-75=60), but again that didn't work. sorry for asking again, If your not busy could you please reply. thankyou

The bearing from A back toward C is 180° different from the bearing CA: bearing AC = 210° - 180° = 30°. Given bearing from A to B is 75°. Angle CAB is the difference of those two bearings.

 

[neelam]

confused, sorry

The bearing from A back toward C is 180° different from the bearing CA: bearing AC = 210° - 180° = 30°. Given bearing from A to B is 75°. Angle CAB is the difference of those two bearings.[/QUOTE)

I'm sorry, I am totally lost. I don't understand why the bearing from A back toward C is 180 degrees. I thought it would have been less. Also to the reply before that I understand where the axis come from but I don't get how you get 30. so I think I'm doing something wrong. so I have drawn a diagram of what I think it should look like. sorry again,

 

[srmichael]

The bearing from A back toward C is 180° different from the bearing CA: bearing AC = 210° - 180° = 30°. Given bearing from A to B is 75°. Angle CAB is the difference of those two bearings.[/QUOTE)

I'm sorry, I am totally lost. I don't understand why the bearing from A back toward C is 180 degrees. I thought it would have been less. Also to the reply before that I understand where the axis come from but I don't get how you get 30. so I think I'm doing something wrong. so I have drawn a diagram of what I think it should look like. sorry again,

That blue angle you reference is 30°.

 

[neelam]

Oh right, thank you.

 

[DrPhil]

I'm sorry, I am totally lost. I don't understand why the bearing from A back toward C is 180 degrees. I thought it would have been less. Also to the reply before that I understand where the axis come from but I don't get how you get 30. so I think I'm doing something wrong. so I have drawn a diagram of what I think it should look like. sorry again,

The bearing from A back toward C is 180° different from the bearing CA:

C to A bearing = 210°
A to C bearing = 210° - 180° = 30°.

Your picture is very good.